Combining Short Multiplication Formulas Practice Problems

To solve the given equation, follow these steps:

• Step 1: Expand $(x - 4)^2$ using the formula $(a - b)^2 = a^2 - 2ab + b^2$.

Thus, $(x - 4)^2 = x^2 - 8x + 16$.

• Step 2: Substitute the expanded form into the equation:

$x^2 - 8x + 16 + 3x^2 = -16x + 12$.

• Step 3: Combine like terms on the left-hand side.

This gives $4x^2 - 8x + 16 = -16x + 12$.

• Step 4: Rearrange the equation to set it to zero.

Bring all terms to one side: $4x^2 - 8x + 16 + 16x - 12 = 0$.

Combine and simplify the terms: $4x^2 + 8x + 4 = 0$.

• Step 5: Simplify the equation by dividing each term by 4.

It becomes $x^2 + 2x + 1 = 0$.

• Step 6: Recognize the equation as a perfect square trinomial.

$(x + 1)^2 = 0$.

• Step 7: Solve by taking the square root of both sides.

The solution is $x + 1 = 0$, therefore $x = -1$.

In conclusion, the solution to the equation is $x = -1$.

We will solve the equation $(x+2)^2 = (2x+3)^2$ by expanding and simplifying both sides:

Step 1: Expand both sides of the equation:
Left side: $(x+2)^2 = x^2 + 4x + 4$
Right side: $(2x+3)^2 = 4x^2 + 12x + 9$

Step 2: Set the expanded forms equal to each other:
$x^2 + 4x + 4 = 4x^2 + 12x + 9$

Step 3: Rearrange to form a standard quadratic equation:
Subtract $x^2 + 4x + 4$ from both sides:
$0 = 3x^2 + 8x + 5$

Step 4: Rearrange to get:
$3x^2 + 8x + 5 = 0$

Step 5: Solve using the quadratic formula:
Using $a = 3$, $b = 8$, $c = 5$:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
$x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 3 \cdot 5}}{2 \cdot 3}$
$x = \frac{-8 \pm \sqrt{64 - 60}}{6}$
$x = \frac{-8 \pm \sqrt{4}}{6}$
$x = \frac{-8 \pm 2}{6}$

Step 6: Calculate the solutions:
$x_1 = \frac{-8 + 2}{6} = \frac{-6}{6} = -1$
$x_2 = \frac{-8 - 2}{6} = \frac{-10}{6} = -\frac{5}{3}$

Verify in the original equation to assure correctness. Hence, both solutions are valid.

Therefore, the solutions are $x_1 = -1$ and $x_2 = -\frac{5}{3}$, which matches choice 3.

To solve the equation $7x + 1 + (2x + 3)^2 = (4x + 2)^2$, we follow these steps:

• Step 1: Expand both sides using the square of a binomial formula.
• Step 2: Simplify the equation to form a standard quadratic equation.
• Step 3: Use the quadratic formula to find the roots of the equation.

Step 1: Expand the squares.

The left side: $(2x + 3)^2 = 4x^2 + 12x + 9$.

The right side: $(4x + 2)^2 = 16x^2 + 16x + 4$.

Step 2: Substitute back into the original equation and simplify:

$7x + 1 + 4x^2 + 12x + 9 = 16x^2 + 16x + 4$.

Combine like terms:

$4x^2 + 19x + 10 = 16x^2 + 16x + 4$.

Step 3: Move all terms to one side:

$4x^2 + 19x + 10 - 16x^2 - 16x - 4 = 0$.

Which simplifies to:

$-12x^2 + 3x + 6 = 0$.

Step 4: Divide by -3 to simplify:

$4x^2 - x - 2 = 0$.

Step 5: Use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 4$, $b = -1$, $c = -2$.

Calculate the discriminant:

$b^2 - 4ac = (-1)^2 - 4 \cdot 4 \cdot (-2) = 1 + 32 = 33$.

Calculate the roots:

$x = \frac{1 \pm \sqrt{33}}{8}$.

Therefore, the solution to the problem is $x = \frac{1 \pm \sqrt{33}}{8}$.

To solve the equation $(x+3)^2 = 2x + 5$, we proceed as follows:

• Step 1: Expand the left side. Using the identity $(a+b)^2 = a^2 + 2ab + b^2$, we find:
  $(x+3)^2 = x^2 + 6x + 9$.

• Step 2: Set the equation to zero by moving all terms to one side:
  $x^2 + 6x + 9 = 2x + 5$
  Subtract $2x + 5$ from both sides:
  $x^2 + 6x + 9 - 2x - 5 = 0$
  This simplifies to:
  $x^2 + 4x + 4 = 0$.

• Step 3: Solve the quadratic equation $x^2 + 4x + 4 = 0$. Notice this can be factored as:
  $(x+2)^2 = 0$.

• Step 4: Solve for $x$ by setting the factor equal to zero:
  $x+2 = 0$.
  Thus, $x = -2$.

Therefore, the solution to the equation $(x+3)^2 = 2x + 5$ is $x = -2$.

The given equation is:

$2x^2 - 2x = (x+1)^2$

Step 1: Expand the right-hand side.

$(x+1)^2 = x^2 + 2x + 1$

Step 2: Write the full equation with the expanded form.

$2x^2 - 2x = x^2 + 2x + 1$

Step 3: Bring all terms to one side of the equation to set it to zero.

$2x^2 - 2x - x^2 - 2x - 1 = 0$

Step 4: Simplify the equation.

$x^2 - 4x - 1 = 0$

Step 5: Identify coefficients for the quadratic formula.

Here, $a = 1$, $b = -4$, $c = -1$.

Step 6: Apply the quadratic formula.

$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1}$

$x = \frac{4 \pm \sqrt{16 + 4}}{2}$

$x = \frac{4 \pm \sqrt{20}}{2}$

$x = \frac{4 \pm 2\sqrt{5}}{2}$

$x = 2 \pm \sqrt{5}$

Therefore, the solutions are $x = 2 + \sqrt{5}$ and $x = 2 - \sqrt{5}$.

These solutions correspond to choice (4): Answers a + b